[docs]classPolyhedron(Basic):""" Represents the polyhedral symmetry group (PSG). The PSG is one of the symmetry groups of the Platonic solids. There are three polyhedral groups: the tetrahedral group of order 12, the octahedral group of order 24, and the icosahedral group of order 60. All doctests have been given in the docstring of the constructor of the object. References ========== http://mathworld.wolfram.com/PolyhedralGroup.html """_edges=Nonedef__new__(cls,corners,faces=[],pgroup=[]):""" The constructor of the Polyhedron group object. It takes up to three parameters: the corners, faces, and allowed transformations. The corners/vertices are entered as a list of arbitrary expressions that are used to identify each vertex. The faces are entered as a list of tuples of indices; a tuple of indices identifies the vertices which define the face. They should be entered in a cw or ccw order; they will be standardized by reversal and rotation to be give the lowest lexical ordering. If no faces are given then no edges will be computed. >>> from sympy.combinatorics.polyhedron import Polyhedron >>> Polyhedron(list('abc'), [(1, 2, 0)]).faces {(0, 1, 2)} >>> Polyhedron(list('abc'), [(1, 0, 2)]).faces {(0, 1, 2)} The allowed transformations are entered as allowable permutations of the vertices for the polyhedron. Instance of Permutations (as with faces) should refer to the supplied vertices by index. These permutation are stored as a PermutationGroup. Examples ======== >>> from sympy.combinatorics.permutations import Permutation >>> Permutation.print_cyclic = False >>> from sympy.abc import w, x, y, z Here we construct the Polyhedron object for a tetrahedron. >>> corners = [w, x, y, z] >>> faces = [(0,1,2), (0,2,3), (0,3,1), (1,2,3)] Next, allowed transformations of the polyhedron must be given. This is given as permutations of vertices. Although the vertices of a tetrahedron can be numbered in 24 (4!) different ways, there are only 12 different orientations for a physical tetrahedron. The following permutations, applied once or twice, will generate all 12 of the orientations. (The identity permutation, Permutation(range(4)), is not included since it does not change the orientation of the vertices.) >>> pgroup = [Permutation([[0,1,2], [3]]), \ Permutation([[0,1,3], [2]]), \ Permutation([[0,2,3], [1]]), \ Permutation([[1,2,3], [0]]), \ Permutation([[0,1], [2,3]]), \ Permutation([[0,2], [1,3]]), \ Permutation([[0,3], [1,2]])] The Polyhedron is now constructed and demonstrated: >>> tetra = Polyhedron(corners, faces, pgroup) >>> tetra.size 4 >>> tetra.edges {(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)} >>> tetra.corners (w, x, y, z) It can be rotated with an arbitrary permutation of vertices, e.g. the following permutation is not in the pgroup: >>> tetra.rotate(Permutation([0, 1, 3, 2])) >>> tetra.corners (w, x, z, y) An allowed permutation of the vertices can be constructed by repeatedly applying permutations from the pgroup to the vertices. Here is a demonstration that applying p and p**2 for every p in pgroup generates all the orientations of a tetrahedron and no others: >>> all = ( (w, x, y, z), \ (x, y, w, z), \ (y, w, x, z), \ (w, z, x, y), \ (z, w, y, x), \ (w, y, z, x), \ (y, z, w, x), \ (x, z, y, w), \ (z, y, x, w), \ (y, x, z, w), \ (x, w, z, y), \ (z, x, w, y) ) >>> got = [] >>> for p in (pgroup + [p**2 for p in pgroup]): ... h = Polyhedron(corners) ... h.rotate(p) ... got.append(h.corners) ... >>> set(got) == set(all) True The make_perm method of a PermutationGroup will randomly pick permutations, multiply them together, and return the permutation that can be applied to the polyhedron to give the orientation produced by those individual permutations. Here, 3 permutations are used: >>> tetra.pgroup.make_perm(3) # doctest: +SKIP Permutation([0, 3, 1, 2]) To select the permutations that should be used, supply a list of indices to the permutations in pgroup in the order they should be applied: >>> use = [0, 0, 2] >>> p002 = tetra.pgroup.make_perm(3, use) >>> p002 Permutation([1, 0, 3, 2]) Apply them one at a time: >>> tetra.reset() >>> for i in use: ... tetra.rotate(pgroup[i]) ... >>> tetra.vertices (x, w, z, y) >>> sequentially = tetra.vertices Apply the composite permutation: >>> tetra.reset() >>> tetra.rotate(p002) >>> tetra.corners (x, w, z, y) >>> tetra.corners in all and tetra.corners == sequentially True Notes ===== Defining permutation groups --------------------------- It is not necessary to enter any permutations, nor is necessary to enter a complete set of transforations. In fact, for a polyhedron, all configurations can be constructed from just two permutations. For example, the orientations of a tetrahedron can be generated from an axis passing through a vertex and face and another axis passing through a different vertex or from an axis passing through the midpoints of two edges opposite of each other. For simplicity of presentation, consider a square -- not a cube -- with vertices 1, 2, 3, and 4: 1-----2 We could think of axes of rotation being: | | 1) through the face | | 2) from midpoint 1-2 to 3-4 or 1-3 to 2-4 3-----4 3) lines 1-4 or 2-3 To determine how to write the permutations, imagine 4 cameras, one at each corner, labeled A-D: A B A B 1-----2 1-----3 vertex index: | | | | 1 0 | | | | 2 1 3-----4 2-----4 3 2 C D C D 4 3 original after rotation along 1-4 A diagonal and a face axis will be chosen for the "permutation group" from which any orientation can be constructed. >>> pgroup = [] Imagine a clockwise rotation when viewing 1-4 from camera A. The new orientation is (in camera-order): 1, 3, 2, 4 so the permutation is given using the *indices* of the vertices as: >>> pgroup.append(Permutation((0, 2, 1, 3))) Now imagine rotating clockwise when looking down an axis entering the center of the square as viewed. The new camera-order would be 3, 1, 4, 2 so the permutation is (using indices): >>> pgroup.append(Permutation((2, 0, 3, 1))) The square can now be constructed: ** use real-world labels for the vertices, entering them in camera order ** for the faces we use zero-based indices of the vertices in *edge-order* as the face is traversed; neither the direction nor the starting point matter -- the faces are only used to define edges (if so desired). >>> square = Polyhedron((1, 2, 3, 4), [(0, 1, 3, 2)], pgroup) To rotate the square with a single permutation we can do: >>> square.rotate(square.pgroup[0]); square.corners (1, 3, 2, 4) To use more than one permutation (or to use one permutation more than once) it is more convenient to use the make_perm method: >>> p011 = square.pgroup.make_perm([0,1,1]) # diag flip + 2 rotations >>> square.reset() # return to initial orientation >>> square.rotate(p011); square.corners (4, 2, 3, 1) Thinking outside the box ------------------------ Although the Polyhedron object has a direct physical meaning, it actually has broader application. In the most general sense it is just a decorated PermutationGroup, allowing one to connect the permutations to something physical. For example, a Rubik's cube is not a proper polyhedron, but the Polyhedron class can be used to represent it in a way that helps to visualize the Rubik's cube. >>> from sympy.utilities.iterables import flatten, unflatten >>> from sympy import symbols >>> from sympy.combinatorics import RubikGroup >>> facelets = flatten([symbols(s+'1:5') for s in 'UFRBLD']) >>> def show(): ... pairs = unflatten(r2.corners, 2) ... print(pairs[::2]) ... print(pairs[1::2]) ... >>> r2 = Polyhedron(facelets, pgroup=RubikGroup(2)) >>> show() [(U1, U2), (F1, F2), (R1, R2), (B1, B2), (L1, L2), (D1, D2)] [(U3, U4), (F3, F4), (R3, R4), (B3, B4), (L3, L4), (D3, D4)] >>> r2.rotate(0) # cw rotation of F >>> show() [(U1, U2), (F3, F1), (U3, R2), (B1, B2), (L1, D1), (R3, R1)] [(L4, L2), (F4, F2), (U4, R4), (B3, B4), (L3, D2), (D3, D4)] Predefined Polyhedra ==================== For convenience, the vertices and faces are defined for the following standard solids along with a permutation group for transformations. When the polyhedron is oriented as indicated below, the vertices in a given horizontal plane are numbered in ccw direction, starting from the vertex that will give the lowest indices in a given face. (In the net of the vertices, indices preceded by "-" indicate replication of the lhs index in the net.) tetrahedron, tetrahedron_faces ------------------------------ 4 vertices (vertex up) net: 0 0-0 1 2 3-1 4 faces: (0,1,2) (0,2,3) (0,3,1) (1,2,3) cube, cube_faces ---------------- 8 vertices (face up) net: 0 1 2 3-0 4 5 6 7-4 6 faces: (0,1,2,3) (0,1,5,4) (1,2,6,5) (2,3,7,6) (0,3,7,4) (4,5,6,7) octahedron, octahedron_faces ---------------------------- 6 vertices (vertex up) net: 0 0 0-0 1 2 3 4-1 5 5 5-5 8 faces: (0,1,2) (0,2,3) (0,3,4) (0,1,4) (1,2,5) (2,3,5) (3,4,5) (1,4,5) dodecahedron, dodecahedron_faces -------------------------------- 20 vertices (vertex up) net: 0 1 2 3 4 -0 5 6 7 8 9 -5 14 10 11 12 13-14 15 16 17 18 19-15 12 faces: (0,1,2,3,4) (0,1,6,10,5) (1,2,7,11,6) (2,3,8,12,7) (3,4,9,13,8) (0,4,9,14,5) (5,10,16,15,14) ( 6,10,16,17,11) (7,11,17,18,12) (8,12,18,19,13) (9,13,19,15,14) (15,16,17,18,19) icosahedron, icosahedron_faces ------------------------------ 12 vertices (face up) net: 0 0 0 0 -0 1 2 3 4 5 -1 6 7 8 9 10 -6 11 11 11 11 -11 20 faces: (0,1,2) (0,2,3) (0,3,4) (0,4,5) (0,1,5) (1,2,6) (2,3,7) (3,4,8) (4,5,9) (1,5,10) (2,6,7) (3,7,8) (4,8,9) (5,9,10) (1,6,10) (6,7,11,) (7,8,11) (8,9,11) (9,10,11) (6,10,11) >>> from sympy.combinatorics.polyhedron import cube >>> cube.edges {(0, 1), (0, 3), (0, 4), '...', (4, 7), (5, 6), (6, 7)} If you want to use letters or other names for the corners you can still use the pre-calculated faces: >>> corners = list('abcdefgh') >>> Polyhedron(corners, cube.faces).corners (a, b, c, d, e, f, g, h) References ========== [1] www.ocf.berkeley.edu/~wwu/articles/platonicsolids.pdf """faces=[minlex(f,directed=False,is_set=True)forfinfaces]corners,faces,pgroup=args= \
[Tuple(*a)forain(corners,faces,pgroup)]obj=Basic.__new__(cls,*args)obj._corners=tuple(corners)# in order givenobj._faces=FiniteSet(faces)ifpgroupandpgroup[0].size!=len(corners):raiseValueError("Permutation size unequal to number of corners.")# use the identity permutation if none are givenobj._pgroup=PermutationGroup((pgroupor[Perm(range(len(corners)))]))returnobj@property

[docs]defcyclic_form(self):"""Return the indices of the corners in cyclic notation. The indices are given relative to the original position of corners. See Also ======== corners, array_form """returnPerm._af_new(self.array_form).cyclic_form

@property

[docs]defsize(self):""" Get the number of corners of the Polyhedron. """returnlen(self._corners)

@property

[docs]deffaces(self):""" Get the faces of the Polyhedron. """returnself._faces

@property

[docs]defpgroup(self):""" Get the permutations of the Polyhedron. """returnself._pgroup

[docs]defrotate(self,perm):""" Apply a permutation to the polyhedron *in place*. The permutation may be given as a Permutation instance or an integer indicating which permutation from pgroup of the Polyhedron should be applied. This is an operation that is analogous to rotation about an axis by a fixed increment. Notes ===== When a Permutation is applied, no check is done to see if that is a valid permutation for the Polyhedron. For example, a cube could be given a permutation which effectively swaps only 2 vertices. A valid permutation (that rotates the object in a physical way) will be obtained if one only uses permutations from the ``pgroup`` of the Polyhedron. On the other hand, allowing arbitrary rotations (applications of permutations) gives a way to follow named elements rather than indices since Polyhedron allows vertices to be named while Permutation works only with indices. Examples ======== >>> from sympy.combinatorics import Polyhedron, Permutation >>> from sympy.combinatorics.polyhedron import cube >>> cube.corners (0, 1, 2, 3, 4, 5, 6, 7) >>> cube.rotate(0) >>> cube.corners (1, 2, 3, 0, 5, 6, 7, 4) A non-physical "rotation" that is not prohibited by this method: >>> cube.reset() >>> cube.rotate(Permutation([[1,2]], size=8)) >>> cube.corners (0, 2, 1, 3, 4, 5, 6, 7) Polyhedron can be used to follow elements of set that are identified by letters instead of integers: >>> shadow = h5 = Polyhedron(list('abcde')) >>> p = Permutation([3, 0, 1, 2, 4]) >>> h5.rotate(p) >>> h5.corners (d, a, b, c, e) >>> _ == shadow.corners True >>> copy = h5.copy() >>> h5.rotate(p) >>> h5.corners == copy.corners False """ifnotisinstance(perm,Perm):perm=self.pgroup[perm]# and we know it's validelse:ifperm.size!=self.size:raiseValueError('Polyhedron and Permutation sizes differ.')a=perm.array_formcorners=[self.corners[a[i]]foriinrange(len(self.corners))]self._corners=tuple(corners)

def_pgroup_calcs():"""Return the permutation groups for each of the polyhedra and the face definitions: tetrahedron, cube, octahedron, dodecahedron, icosahedron, tetrahedron_faces, cube_faces, octahedron_faces, dodecahedron_faces, icosahedron_faces (This author didn't find and didn't know of a better way to do it though there likely is such a way.) Although only 2 permutations are needed for a polyhedron in order to generate all the possible orientations, a group of permutations is provided instead. A set of permutations is called a "group" if:: a*b = c (for any pair of permutations in the group, a and b, their product, c, is in the group) a*(b*c) = (a*b)*c (for any 3 permutations in the group associativity holds) there is an identity permutation, I, such that I*a = a*I for all elements in the group a*b = I (the inverse of each permutation is also in the group) None of the polyhedron groups defined follow these definitions of a group. Instead, they are selected to contain those permutations whose powers alone will construct all orientations of the polyhedron, i.e. for permutations ``a``, ``b``, etc... in the group, ``a, a**2, ..., a**o_a``, ``b, b**2, ..., b**o_b``, etc... (where ``o_i`` is the order of permutation ``i``) generate all permutations of the polyhedron instead of mixed products like ``a*b``, ``a*b**2``, etc.... Note that for a polyhedron with n vertices, the valid permutations of the vertices exclude those that do not maintain its faces. e.g. the permutation BCDE of a square's four corners, ABCD, is a valid permutation while CBDE is not (because this would twist the square). Examples ======== The is_group checks for: closure, the presence of the Identity permutation, and the presence of the inverse for each of the elements in the group. This confirms that none of the polyhedra are true groups: >>> from sympy.combinatorics.polyhedron import ( ... tetrahedron, cube, octahedron, dodecahedron, icosahedron) ... >>> polyhedra = (tetrahedron, cube, octahedron, dodecahedron, icosahedron) >>> [h.pgroup.is_group() for h in polyhedra] ... [False, False, False, False, False] Although tests in polyhedron's test suite check that powers of the permutations in the groups generate all permutations of the vertices of the polyhedron, here we also demonstrate the powers of the given permutations create a complete group for the tetrahedron: >>> from sympy.combinatorics import Permutation, PermutationGroup >>> for h in polyhedra[:1]: ... G = h.pgroup ... perms = set() ... for g in G: ... for e in range(g.order()): ... p = tuple((g**e).array_form) ... perms.add(p) ... ... perms = [Permutation(p) for p in perms] ... assert PermutationGroup(perms).is_group() In addition to doing the above, the tests in the suite confirm that the faces are all present after the application of each permutation. References ========== http://dogschool.tripod.com/trianglegroup.html """def_pgroup_of_double(polyh,ordered_faces,pgroup):n=len(ordered_faces[0])# the vertices of the double which sits inside a give polyhedron# can be found by tracking the faces of the outer polyhedron.# A map between face and the vertex of the double is made so that# after rotation the position of the vertices can be locatedfmap=dict(zip(ordered_faces,range(len(ordered_faces))))flat_faces=flatten(ordered_faces)new_pgroup=[]fori,pinenumerate(pgroup):h=polyh.copy()h.rotate(p)c=h.corners# reorder corners in the order they should appear when# enumerating the facesreorder=unflatten([c[j]forjinflat_faces],n)# make them canonicalreorder=[tuple(map(as_int,minlex(f,directed=False,is_set=True)))forfinreorder]# map face to vertex: the resulting list of vertices are the# permutation that we seek for the doublenew_pgroup.append(Perm([fmap[f]forfinreorder]))returnnew_pgrouptetrahedron_faces=[(0,1,2),(0,2,3),(0,3,1),# upper 3(1,2,3),# bottom]# cw from top#_t_pgroup=[Perm([[1,2,3],[0]]),# cw from topPerm([[0,1,2],[3]]),# cw from front facePerm([[0,3,2],[1]]),# cw from back right facePerm([[0,3,1],[2]]),# cw from back left facePerm([[0,1],[2,3]]),# through front left edgePerm([[0,2],[1,3]]),# through front right edgePerm([[0,3],[1,2]]),# through back edge]tetrahedron=Polyhedron(range(4),tetrahedron_faces,_t_pgroup)cube_faces=[(0,1,2,3),# upper(0,1,5,4),(1,2,6,5),(2,3,7,6),(0,3,7,4),# middle 4(4,5,6,7),# lower]# U, D, F, B, L, R = up, down, front, back, left, right_c_pgroup=[Perm(p)forpin[[1,2,3,0,5,6,7,4],# cw from top, U[4,0,3,7,5,1,2,6],# cw from F face[4,5,1,0,7,6,2,3],# cw from R face[1,0,4,5,2,3,7,6],# cw through UF edge[6,2,1,5,7,3,0,4],# cw through UR edge[6,7,3,2,5,4,0,1],# cw through UB edge[3,7,4,0,2,6,5,1],# cw through UL edge[4,7,6,5,0,3,2,1],# cw through FL edge[6,5,4,7,2,1,0,3],# cw through FR edge[0,3,7,4,1,2,6,5],# cw through UFL vertex[5,1,0,4,6,2,3,7],# cw through UFR vertex[5,6,2,1,4,7,3,0],# cw through UBR vertex[7,4,0,3,6,5,1,2],# cw through UBL]]cube=Polyhedron(range(8),cube_faces,_c_pgroup)octahedron_faces=[(0,1,2),(0,2,3),(0,3,4),(0,1,4),# top 4(1,2,5),(2,3,5),(3,4,5),(1,4,5),# bottom 4]octahedron=Polyhedron(range(6),octahedron_faces,_pgroup_of_double(cube,cube_faces,_c_pgroup))dodecahedron_faces=[(0,1,2,3,4),# top(0,1,6,10,5),(1,2,7,11,6),(2,3,8,12,7),# upper 5(3,4,9,13,8),(0,4,9,14,5),(5,10,16,15,14),(6,10,16,17,11),(7,11,17,18,12),# lower 5(8,12,18,19,13),(9,13,19,15,14),(15,16,17,18,19)# bottom]def_string_to_perm(s):rv=[Perm(range(20))]p=Noneforsiins:ifsinotin'01':count=int(si)-1else:count=1ifsi=='0':p=_f0elifsi=='1':p=_f1rv.extend([p]*count)returnPerm.rmul(*rv)# top face cw_f0=Perm([1,2,3,4,0,6,7,8,9,5,11,12,13,14,10,16,17,18,19,15])# front face cw_f1=Perm([5,0,4,9,14,10,1,3,13,15,6,2,8,19,16,17,11,7,12,18])# the strings below, like 0104 are shorthand for F0*F1*F0**4 and are# the remaining 4 face rotations, 15 edge permutations, and the# 10 vertex rotations._dodeca_pgroup=[_f0,_f1]+[_string_to_perm(s)forsin''' 0104 140 014 0410 010 1403 03104 04103 102 120 1304 01303 021302 03130 0412041 041204103 04120410 041204104 041204102 10 01 1402 0140 04102 0412 1204 1302 0130 03120'''.strip().split()]dodecahedron=Polyhedron(range(20),dodecahedron_faces,_dodeca_pgroup)icosahedron_faces=[[0,1,2],[0,2,3],[0,3,4],[0,4,5],[0,1,5],[1,6,7],[1,2,7],[2,7,8],[2,3,8],[3,8,9],[3,4,9],[4,9,10],[4,5,10],[5,6,10],[1,5,6],[6,7,11],[7,8,11],[8,9,11],[9,10,11],[6,10,11]]icosahedron=Polyhedron(range(12),icosahedron_faces,_pgroup_of_double(dodecahedron,dodecahedron_faces,_dodeca_pgroup))return(tetrahedron,cube,octahedron,dodecahedron,icosahedron,tetrahedron_faces,cube_faces,octahedron_faces,dodecahedron_faces,icosahedron_faces)(tetrahedron,cube,octahedron,dodecahedron,icosahedron,tetrahedron_faces,cube_faces,octahedron_faces,dodecahedron_faces,icosahedron_faces)=_pgroup_calcs()